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Topic:
Contour integral
Replies:
3
Last Post:
May 26, 2012 11:23 AM




Re: Contour integral
Posted:
May 26, 2012 12:53 AM


On 25May2012, Robin Chapman <R.J.Chapman@ex.ac.uk> wrote in message <jpnjar$lrp$1@dontemail.me>:
> > I(a,b) = Int_{inf}^{+inf} exp[a^2*(x+b)^2] dx > > > > where a and b are complex and Re(a^2)> 0 (so the integral > > converges). > > > > The text claims without proof that: > > > > The method of residues enables us to show that this integral > > does not depend on b: > > > > I(a,b) = I(a,0) > > Integrate exp(a^2 z^2) over rectangular contour: vertices +R, > +R + i Im(b). > > > and that, when the condition pi/4< Arg(a)< +pi/4 is > > fulfilled (which is always possible if Re(a^2)> 0), I(a,0) > > is given by: > > > > I(a,0) = I(1,0)/a = sqrt(pi)/a > > Integrate exp(z^2) over "sliceofpie" contour: centre 0 > vertices at aR and aR.
Thanks, Robin. And may I say it's nice to see you here again in <sci.math>. I've noticed you've poked your nose in once or twice recently  hope you stick around awhile.
 Jim Heckman



